Trying to prove a consequence of harmonic gauge in GR

[hideelo]

So, I am following the PI lecture series by Neil Turok. He starts with the following description of harmonic gauge condition

$$g^{\mu \nu}\Gamma^{\lambda}_{\mu \nu}=0$$
He then claims that for linearized gravity (weak field) i.e.
$$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu} $$ with $$ |h_{\mu \nu}| <<1 $$ that harmonic gauge is equivalent to the condition that $$h^{\lambda}_{\nu , \lambda} - \frac{1}{2} h^{\lambda}_{\lambda , \nu} = 0$$

My problem is in proving this is how do I prove this, I've been trying for a few days now with no luck, I really need some pointers. I tried using the definition of the Christoffel symbol to work it out and I think I'm going nowhere. Any help would be appreciated.

TIA

 

[WannabeNewton]

If you don't show any of your work, how are we supposed to help? We don't know where exactly you got stuck. For example, did you make sure to work to only first order in the perturbation when computing the connection coefficients?

 

[samalkhaiat]

So, I am following the PI lecture series by Neil Turok. He starts with the following description of harmonic gauge condition

$$g^{\mu \nu}\Gamma^{\lambda}_{\mu \nu}=0$$
He then claims that for linearized gravity (weak field) i.e.
$$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu} $$ with $$ |h_{\mu \nu}| <<1 $$ that harmonic gauge is equivalent to the condition that $$h^{\lambda}_{\nu , \lambda} - \frac{1}{2} h^{\lambda}_{\lambda , \nu} = 0$$

My problem is in proving this is how do I prove this, I've been trying for a few days now with no luck, I really need some pointers. I tried using the definition of the Christoffel symbol to work it out and I think I'm going nowhere. Any help would be appreciated.

TIA

Use [tex]\nabla_{\mu}(\sqrt{-g}g^{\mu\nu}) = 0,[/tex] and rewrite the Harmonic condition in the form [tex]\partial_{\mu}(\sqrt{-g}g^{\mu\nu}) = 0 .[/tex] Expand and contract with [itex]g_{\nu\tau}[/itex] to obtain [tex]\frac{1}{\sqrt{-g}}\partial_{\tau}\sqrt{-g} + g_{\nu\tau}\partial_{\mu}g^{\mu\nu} = 0 .[/tex] This is the same as [tex]\frac{1}{2} g^{\rho\sigma}\partial_{\tau}g_{\rho\sigma} + g_{\nu\tau}\partial_{\mu}g^{\mu\nu} = 0 .[/tex] Now use [itex]g_{\mu\nu} = \eta_{\mu\nu} + \epsilon \ h_{\mu\nu}[/itex] and ignore the [itex]\mathcal{O}(\epsilon^{2})[/itex] terms.

 

[hideelo]

If you don't show any of your work, how are we supposed to help? We don't know where exactly you got stuck. For example, did you make sure to work to only first order in the perturbation when computing the connection coefficients?

So what I have so far is $$0 = g^{\mu \nu} \Gamma^{\lambda}_{\mu \nu} = \frac{1}{2}g^{\mu \nu}g^{\tau \lambda}(g_{\mu \tau ,\nu} + g_{\nu \tau , \mu} -g_{\mu \nu , \tau}) = \frac{1}{2} g^{\tau \lambda}(g_{\mu \tau} \, ^{, \mu} + g_{\nu \tau} \, ^{, \nu}) - \frac{1}{2}g^{\mu \nu}g_{\mu \nu} \, ^{, \lambda} = g^{\tau \lambda} g_{\mu \tau} \, ^{, \mu} - \frac{1}{2}g^{\mu \nu}g_{\mu \nu} \, ^{, \lambda}$$

Now I do the following
$$g_{\mu \tau} \, ^{,\mu} = \partial_\mu g_{\mu \tau} = \partial_\mu( \eta_{\mu \tau} + h_{\mu \tau}) = \partial_\mu h_{\mu \tau} = h_{\mu \tau} \, ^{,\mu}$$ doing this for both terms gives me $$g^{\tau \lambda} h_{\mu \tau} \, ^{, \mu} - \frac{1}{2}g^{\mu \nu} h_{\mu \nu} \, ^{, \lambda} = 0$$

This is what I got so far.

Edit: What it looks like I should do is use the metric tensor to raise the indices on the derivatives of h in the final line, but I don't think I can do that because the derivatives of h aren't tensors so it would be like trying to raise indices on the christoffel symbol, am I wrong?

 

[hideelo]

bump?

 

[samalkhaiat]

bump?

Why? Didn’t I show you exactly how to solve your problem? Do you even know why the de Donder condition ,[itex]g^{\mu\nu}\Gamma_{\mu\nu}^{\rho}=0[/itex], is called harmonic coordinate condition?

 

[hideelo]

Why? Didn’t I show you exactly how to solve your problem? Do you even know why the de Donder condition ,[itex]g^{\mu\nu}\Gamma_{\mu\nu}^{\rho}=0[/itex], is called harmonic coordinate condition?

First of all, thanks for answering.

As I understand it, the reason why its called harmonic gauge is because under this condition the coordinate functions are harmonic i.e. $$\Box x^\mu = 0$$

the bump was in response to this:

If you don't show any of your work, how are we supposed to help? We don't know where exactly you got stuck. For example, did you make sure to work to only first order in the perturbation when computing the connection coefficients?

who asked me to show the work I had done, but then didnt respond, and I wanted to know if I had gone in the totally wrong direction or if I was almost there.

 

[stedwards]

Of course. The de Dongle condition. Everybody knows it..